(Asian-Pacific Olympiad $2017$). Let $a, b, c$ be positive rational numbers with $abc = 1$. Suppose there exist positive integers $x, y, z$ for which $a^x +b^y +c^z$ is an integer. Prove that when $a, b, c$ are written as a fractions in lowest terms, the numerators are perfect powers.
Solution from book OTIS excerpts:
It is suffcient to prove the following claim.
Claim: Let $p$ be a prime. If $\nu_{p}(a) > 0$ then $\nu_{p}(a)$ is divisible by $\frac{y+z}{\gcd(y,z)}$ .
Proof.
Note $\nu_{p}(a) + \nu_{p}(b) + \nu_{p}(c) = 0$.
WLOG assume $\nu_{p}(c) < 0$ (hence $\nu_{p}(c^z) < 0)$.
Then since $\nu_{p}(a^x + b^y + c^z) \geq 0$ and $\nu_{p}(a^x) > 0$, we must have $$\nu_{p}(b^y) = \nu_{p}(c^z) \Rightarrow y\nu_{p}(b) = z\nu_{p}p(c)\,.$$
Thus we may set $\nu_{p}(b) = -z'k$ and $\nu_{p}(c) = -y'k$ where $y' =\frac{y}{\gcd(y,z)}$ and $z' =\frac{z}{\gcd(y,z)}$ . Then $$\nu_{p}(a) =-\nu_{p}(b) -\nu_{p}(c) = k \cdot (y' + z')$$ as needed.
Therefore the numerator of a is a perfect $\frac{y+z}{\gcd(y,z)}$th power.
I understood the solution pretty much entirely except for one thing. Why must $\nu_{p}(b^y) = \nu_{p}(c^z)$ be true? Also because i guess that, because i am unable to understand this, i also don't understand the motivation for choosing such $y'$ and $z'$. Why not just just $x$ and $y$ instead? Could someone please provide a simple explanation? Also if you have your own solution i would be very much interested in seeing it and also getting to know the motivation behind your solutions. Thanks.